Whats is the purpose of the

The index of refraction for red light in water is 1.331 and that for blue light is 1.340. If a ray of white light enters the wat

kobusy [5.1K]

Answer:

Angle of refraction for red light is $43.01^{\circ}$

Angle of refraction for blue light is $42.68^{\circ}$

Explanation:

It is given refractive index for red light is $\mu _{red}=1.331$

Refractive index of blue light $\mu _{blue}=1.340$

Angle of incidence $i=65.30^{\circ}$

According to law of refraction $\mu =\frac{sini}{sinr}$

For red light $1.331 =\frac{sin65.30^{\circ}}{sinr}$

$1.331 =\frac{0.908}{sinr}$

$sinr=0.682$

$r=43.01^{\circ}$

Therefore angle of refraction for red light is $43.01^{\circ}$

Similarly for blue light $1.340 =\frac{sin65.30^{\circ}}{sinr}$

$1.340 =\frac{0.908}{sinr}$

$sinr=0.677$

r = $42.68^{\circ}$

Therefore angle of refraction for blue light is $42.68^{\circ}$

6 0

3 years ago

Which is the force that results from use of a machine?

Goshia [24]

Answer:

D. Output force

Explanation:

The input force is the effort used to run the machine and this results in an output force.

3 0

2 years ago

What is the equation for the force of a spring

Komok [63]

Answer:

F=-kx

Explanation:

yan lang po ang alam ko

7 0

3 years ago

Calculate the ratio of the drag force on a jet flying at 950 km/h at an altitude of 10 km to the drag force on a prop-driven tra

Black_prince [1.1K]

Answer:

$\frac{F_1}{F_2}=3.55$

Explanation:

F = Force

C = Drag coefficient equal for both aircrafts

ρ = Density of air

A = Surface area equal for both aircrafts

v = Velocity

$v_2=\frac{2}{5}v_1$

$F_1=\frac{1}{2}\rho_1 CAv_1^2$

$F_2=\frac{1}{2}\rho_2 CAv_2^2\\\Rightarrow F_2=\frac{1}{2}\rho_2 CA\left(\frac{2}{5}v_1\right)^2$

Dividing the above two equations we get

$\frac{F_1}{F_2}=\frac{\frac{1}{2}\rho_1 CAv_1^2}{\frac{1}{2}\rho_2 CA\left(\frac{2}{5}v_1\right)^2}\\\Rightarrow \frac{F_1}{F_2}=\frac{\rho_1}{\rho_2\frac{4}{25}}\\\Rightarrow \frac{F_1}{F_2}=\frac{0.38}{0.67\frac{4}{25}}\\\Rightarrow \frac{F_1}{F_2}=3.55$

The ratio of the drag forces is $\mathbf{\frac{F_1}{F_2}}=\mathbf{3.55}$

5 0

3 years ago

An aircraft is at a standstill. It accelerates to 140kts in 28 seconds. The aircraft weighs 28,000 lbs. How many feet of runway

stiv31 [10]

Answer:

runway use is 3307.8 feet

Explanation:

given data

velocity = 140 kts = 140 × 0.5144 m/s = 72.016 m/s

time = 28 seconds

weight = 28000 lbs

to find out

How many feet of runway was used

solution

we will use here first equation of motion for find acceleration

v = u + at ..............1

here v is velocity given and u is initial velocity that is 0 and a is acceleration and t is time

put here value in equation 1

72.016 = 0 + a(28)

a = 2.572 m/s²

and

now apply third equation of motion

s = ut + 0.5×a×t² .......................2

here s is distance and u is initial speed and t is time and a is acceleration

put here all value in equation 2

s = 0 + 0.5×2.572×28²

s = 1008.24 m = 3307.8 ft

so runway use is 3307.8 feet

5 0

3 years ago